Question

can anyone please solve this fast as possible

Answer 1

${\huge{\pink{\underline{\underline{\purple{\mathbb{\bold{\mathcal{AnSwEr..}}}}}}}}}$

${\large{\blue{\bold{\underline{Given:-}}}}}$

$\sf\implies \: p(x) = {x}^{2} - mx + n$

$\sf\implies \alpha \: \: and \: \: \beta \: zeroes \: of \: p(x)$

${\large{\blue{\bold{\underline{To find:-}}}}}$

▪︎ Value of,

$\sf\implies \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha }$

${\large{\blue{\bold{\underline{ID\:Used:-}}}}}$

$\sf\implies {a}^{2} + {b}^{2} = (a + b) {}^{2} - 2ab.$

${\large{\blue{\bold{\underline{Now,}}}}}$

$\sf \: in \: p(x) \frac{.}{.} - \\ \\\implies \alpha + \beta = \frac{ - b}{a} \\ \\ \implies \alpha + \beta = \frac{ - ( - m)}{1} \\ \\ \implies \alpha + \beta = m \: \: ....eq(1) \\ \\ and \\ \\ \implies \alpha \times b = \frac{c}{a} \\ \\ \implies \alpha \beta = \frac{n}{1} \\ \\ \implies \alpha \beta = n \: \: ...eq(2)$

${\large{\blue{\bold{\underline{Therefore:-}}}}}$

$\sf\implies \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } \\ \\ = \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } \\ \\ = \frac{( \alpha + \beta ) {}^{2} - 2 \alpha \beta }{ \alpha \beta } \\ \\ = \frac{( {m})^{2} - 2(n)}{n} \\ \\ (from \: (1) \: and \: (2)) \\ \\ = \frac{ {m}^{2} - 2n }{n}$

${\small{\green{\boxed{\boxed{\bold{Answer\:=\:m\:square-2n/n.}}}}}}$

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